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Does anyone know of generalizations on what Mumford (Red Book) calls "uniformizing parameters"? For example, given a regular quasi-projective scheme over a finite field $\mathbb{F}$, is there an etale morphism into affine space over $\mathbb{F}$?.

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I'm not an algebraic geometer but at present I'm reading your question as "given a smooth projective variety, find an etale map from it to some affine n-space" and I'm thinking "however can e.g. the projective line map in an etale way to any affine space at all?". Can you explain either how this can be done, or why you're not saying it can be done? – Kevin Buzzard Feb 25 '10 at 0:40
As Scott Carnahan points out in his answer, this can be checked, and the conclusion is that the dimension of everything in sight will have to be zero. (This has nothing to with etaleness, other than that etaleness implies finite fibres: any map from a connected projective variety to an affine scheme will have to be constant, since the coordinates on the affine scheme will have to pull back to constant functions on the connected projective variety.) – Emerton Feb 25 '10 at 3:41
Thank you for the responses. Indeed, the simple example shows that the answer is negative in general; eg, etale morphisms must preserve dimension. I was curious about generalizations of noether normalization. – Ivan Feb 25 '10 at 3:51
The notion of a "uniformizing parameter" requires etale morphisms defined only locally on the source. In particular if $X$ is a regular scheme, and $x \in X$ a point, one can find locally defined functions mapping to a $k(x)$-basis for $m_x/m_x^2$. In characteristic zero, this suffices to give an etale map $U \to \AA^n$ (after potentially further shrinking $U$). In characteristic $p$, the issue is that there is an extra separability hypothesis that is not automatic. This discussion is of a different flavor than the (more global) noether normalization statement.. – Anatoly Preygel Feb 25 '10 at 4:32
Remark: you've edited the question without indicating you edited it, and hence my comment now no longer makes sense. – Kevin Buzzard Feb 25 '10 at 7:43
up vote 3 down vote accepted

This paper by Kedlaya might be what you want, since it contains some rearrangement of the words you used, but I can't really tell from the question. If you want a proper F-scheme to have an etale map to affine space, it has to be a disjoint union of finite F-schemes, and the affine space has to be zero dimensional.

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[EDIT: I re-arrange the answer to make the statements clearer].

The answer to the question is no, for any base field $k$.

First, we can characterize smooth affine varieties $X$ which are étale over an affine space $A=\mathbb A^n_k$.

  • A smooth affine variety $X$ over $k$ is étale over an affine space $A=\mathbb A^n_k$ if and only if the sheaf of differentials $\Omega_{X/k}$ is free of rank $n$, generated by $n$ closed differential forms $df_1, \ldots, df_n$.

Proof: (1) If there exists an étale morphism $f: X\to A={\rm Spec}k[t_1,\ldots, t_n]$, then we have an isomorphism $\Omega_{A/k}\otimes_{O_A} O_X=f^{\star}\Omega_{A/k}\simeq \Omega_{X/k}$. Then take $f_i$ equal to the image of $t_i$ in $O(X)$. (2) If $df_1,\ldots, df_n$ as above exist, we define a morphism $f: X\to A$ to be associated to the moprhism of $k$-algebras $k[t_1,\ldots, t_n]\to O(X)$, $t_i\mapsto f_i$. Then by hypothesis on the $df_i$'s, the canonical morphism $f^{\star}\Omega_{A/k}\to \Omega_{X/k}$ is an isomorphism. Therefore $f$ is étale.

Remark: it is not enough to suppose that $\Omega_{X/k}$ is free to insure that $X$ is étale over $A$. For example, if $X$ is an elliptic curve $E$ minus the origin, then $\Omega_{X/k}$ is free (because $\Omega_{E/k}$ is !), but at least in characteristic $0$, $X$ is not étale over $\mathbb A^1_k$ (if so, it would be finite and étale over $\mathbb A^1$ by considering the extension $E\to \mathbb P^1_k$, but $\mathbb A^1_k$ is simply connected in characteristic $0$). I don't known whether this can happen in positive characteristic. Note that this is already an example (in characteristic 0) of a smooth affine curve which is not étale over $\mathbb A^1$.

Now we construct in any characteristic a smooth affine curve $X$ such that $\Omega_{X/k}$ is not free. By the above, $X$ will not be étale over ${\mathbb A^1}$. Fix a projective smooth connected curve $C$ over $k$ of genus $g>1$. Suppose that for any affine open subset $X$ of $C$, $\Omega_{X/k}$ is free. We want to find a contradiction.

  1. We first reduce to the case $k$ is algebraically closed (for simplicity, this is actually not necessary). I claim that over the algebraic closure $\bar{k}$ of $k$, $\Omega_{X'/\bar{k}}$ is free for any affine open subset $X'$ of $C_{\bar{k}}$. Indeed, $X'$ is defined over a finite extension $K/k$, and the projection $C_{K}\to C$ induces an étale morphism from $X'$ to an affine open subset $X$ of $C$, thus $\Omega_{X'/K}\simeq \Omega_{X/k}\otimes O_{X'}$ is free.

  2. Now we suppose that $k$ is algebraically closed. For any closed point $x\in C$, $\Omega_{X/k}$, where $X=C \setminus {x}$, is free. So $\Omega_{C/k}|_X$ is trivial. This implies that the canonical divisor $K_C$ on $C$ is trivial on $X$, hence linearly equivalent to $(2g-2)[x]$. Fix a point $x_0\in C(k)$. Then the immersion $C\to {\rm Jac}(C)$, $x\mapsto [x-x_0]$, maps $C$ into the $(2g-2)$-torsion part of ${\rm Jac}(C)(k)$ which is finite. Contradicton.

Final remark: if you have a separated scheme $X$ which is étale (or more generally quasi-finite) over any affine scheme $A$, then it has to be quasi-affine. This is because by Zariski's Main Theorem, $X$ is an open subscheme of a scheme $\overline{X}$ which is finite over $A$, hence $\overline{X}$ is affine.

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